Proof
(related to Proposition: How the Boundary Changes the Property of a Set of Being Open)
By hypothesis, $(X,\mathcal O)$ is a topological space and $B\subset X$ its subset.
Ad $(1)$
 Let $x\in B \setminus \delta B$ be an element of the interior.
 Since $x$ is not a boundary point, there is a neighborhood $N_x$ that has no points in common with the complement $B^C$.
 Moreover, we can choose $N_x$ in a way that it has even no points in common with boundary $\delta B.$ Otherwise, if there existed an $y\in N_x \cap \delta B,$ then $N_x$ would be a neighborhood of $y$ and would have no points in common with the complement $B^C.$ But every neighborhood of a boundary point has points in common with the complement.
 Thus, $N_x$ has no boundary point in it.
 Altogether, we have shown that every neighborhood of $x$ is contained in the interior.
 It follows that the interior is open.
Ad $(2)$
 Let $B^C$ be the complement of $B.$
 From the definition of the boundary it follows that $\delta B^C=\delta B$.
 From $(1)$ we can deduce $B^C\setminus \delta B^C)$ is open.
 Therefore $(B^C\setminus \delta B^C)^C$ is closed.
 But $(B^C\setminus \delta B^C)^C=B\cup \delta B,$ which is the closure.
 Therefore, the closure is closed.
Ad $(3)$
 We have $\delta B=(B\cup \delta B)\setminus(B\setminus \delta B).$
 It follows that $X\setminus \delta B=(X\setminus(B\cup\delta B))\cup(B\setminus \delta B),$ which is, according to \( (1)\) and \((2)\), a union of two open sets.
 By the third axiom of topology, $X\setminus \delta B=(\delta B)^C$ is open.
 Therefore, the complement $((\delta B)^C)^C)=\delta B$ is closed, which is the boundary.
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References
Bibliography
 Forster Otto: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984